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Question
In the given figure ABCD is a rhombus with angle A = 67°
If DEC is an equilateral triangle, calculate:
- ∠CBE
- ∠DBE
Solution
Given that the figure ABCD is a rhombus with angle A = 67o
In the rhombus we have
∠A = 67° = ∠C ...[Opposite angles]
∠A + ∠D = 180° ...[Consecutive angles are supplementary.]
⇒ ∠D = 113°
⇒ ∠ABC = 113°
Consider ΔDBC,
DC = CB ...[Sides of rhombus]
So, ΔDBC is an isosceles triangle,
⇒ ∠CDB = ∠CBD
Also,
∠CDB + ∠CBD + ∠BCD = 180°
⇒ 2∠CBD = 113°
⇒ ∠CDB = ∠CBD = 56.5° ...(i)
Consider ΔDCE,
EC = CB
So ΔBCE is an isosceles triangle
⇒ ∠CBE = ∠CEB
Also,
∠CBE + ∠CEB + ∠BCE = 180°
⇒ 2∠CBE = 53°
⇒ ∠CDE = 26.5°
From (i)
∠CBD = 56.5°
⇒ ∠CBE + ∠DBE = 56.5°
⇒ 26.5° + ∠DBE = 56.5°
⇒ ∠DBE = 56.5° - 26.5°
⇒ ∠DBE = 30°
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